# Compund Angle Identities and proof

1. Feb 20, 2014

1. The problem statement, all variables and given/known data
prove using the compound angle identies, proove the following:
$$\frac{sin(A-B)}{cos(A)cos(B)}+\frac{sin(B-C)}{cos(B)cos(C)}+\frac{sin(C-A)}{cos(C)cos(A)}=0$$

2. Relevant equations
n/a

3. The attempt at a solution
I resolved it to
$$\frac{sin(A)cos(B)-cos(A)sin(B)}{cos(A)cos(B)}+\frac{sin(B)cos(C)-cos(B)sin(C)}{cos(B)cos(C)}+\frac{sin(C)cos(A)-cos(C)sin(A)}{cos(C)cos(A)}=0$$

And using wolfram alpha I found that the first term resolves to tan(A)-tan(B) and then I can see how it equals zero as all the tans would cancel out.

But I have no idea how each term simplifies to that.

Any help is really appreciated.

thanks.

2. Feb 20, 2014

### CAF123

After you have expanded the numerators with the addition formulae (which you have done), you can simplify each fraction by taking out a factor.

3. Feb 20, 2014

### Mentallic

In each fraction, you have something of the form

$$\frac{a+b}{c}$$

You should always check if splitting up the fraction into

$$\frac{a}{c}+\frac{b}{c}$$